Do you have a function called
y = (x-3)^2 + 2((x-3)^2 + 2)) ?
then
y = (x-3)^2 + 2(x-3)^2 + 4
= 3(x-3)^2 + 4
to take the inverse,
1. exchange the x and y variables
2. solve the new equation for y
so
x = 3(y-3)^2 + 4
3(y-3)^2 = x-4
(y-3)^2 = (x-4)/3
y-3 = ±√(x-4)/√3
y = 3 ± √(x-4)/√3
what is the inverse of (x-3)^2+2
((x-3)squared plus 2))
2 answers
It looks to me like your wrote the same function twice.
If you meant to write just
y = (x-3)^2 +2, then:
sqrt(y-2) = x-3
x = sqrt(y-2) + 3
and the inverse function is
y = sqrt(x-2) + 3
If you meant to write just
y = (x-3)^2 +2, then:
sqrt(y-2) = x-3
x = sqrt(y-2) + 3
and the inverse function is
y = sqrt(x-2) + 3